Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $k$ be an algebraically closed field. Why there is no isomorphism (as finitely generated $k$-algebras) between the ring $k[x,y]/(x^{2}-y^{5})$ and $k[t]$?

share|improve this question
add comment

6 Answers

up vote 10 down vote accepted

The ring $k[T]$ is a PID, hence integrally closed, whereas $A=k[x,y]/(x^2-y^5)=k[\xi, \eta]$ is not, so these rings are not isomorphic.

To see that $A$ is not integrally closed, just observe that the element $\xi/ \eta^2\in Frac(A)$ is not in $A$ and nevertheless is integral over $A$ since it is a zero of the monic polynomial $X^2-y\in A[X]$.

Generalization
Not only is the hypothesis "$k$ algebraically closed" irrelevant, but the exact same proof works if you only assume that $k$ is a UFD rather than a field, since Gauss tells you that $k[T]$ is then also a UFD and UFD's are integrally closed.

share|improve this answer
1  
Or indeed $k$ is just some integrally closed commutative ring, not even a UFD :-) For example $k$ could be the ring of integers of some number field, and it is then allowed to have class number not equal to 1. –  Alex B. Mar 28 '12 at 8:30
1  
Dear @Alex, you are absolutely right and funnily enough that is the degree of generality I wanted to evoke in my post. However I would then have had to state that if $k$ is integrally closed, then so is $k[T]$. This is true (and proved for example in Lemma 7.31.8 of De Jong and collaborators' online book Stacks Project ) but I don't know an elementary reference, so I downgraded to UFD's. Do you know an elementary reference? –  Georges Elencwajg Mar 28 '12 at 9:13
    
Unfortunately not. Indeed, I don't think I knew any reference until now, I just had this fact stored somewhere in my mind as "folklore wisdom", not having any idea where I had it from. So thank you! –  Alex B. Mar 28 '12 at 9:17
2  
Dear @Alex, then we'll just let interested readers check the reference to the online book I mentioned above. By the way, I think we all have some such "black boxes" stored somewhere in our brains: life is short and mathematics is vast... –  Georges Elencwajg Mar 28 '12 at 9:23
add comment

Let $R = k[x,y]/(x^2-y^5)$. The $k$-homomorphisms $R \to k$ correspond to pairs $a,b \in k$ with $a^2=b^5$. The kernel is a maximal ideal $\mathfrak{m}_{a,b} \subseteq R$, to which we may associate the dimension of the vector space $T^*_{a,b} := \mathfrak{m}_{a,b} / \mathfrak{m}_{a,b}^2$ over $k$ (this is the so-called cotangent space at the rational point $(a,b)$). One computes that $T^*_{0,0}$ has $k$-basis $x,y$, thus is $2$-dimensional. However, every cotangent space of $k[t]$ is $1$-dimensional: For $a \in k$ we $T^*_a=(t-a)k[t]/(t-a)^2 k[t]$, which has basis $t-a$.

enter image description here

This is, in fact, a geometric proof, which can be applied in other situations, too. $\mathrm{Spec}(R)$ is a singular curve (it has a cusp in the point $(0,0)$), whereas $\mathrm{Spec}(k[t]) = \mathbb{A}^1$ is just the affine line, which is a regular curve.

share|improve this answer
add comment

As $k[t]$ is a principal ideal domain, it is integrally closed in its field of fractions. However, $k[x,y]/(x^2-y^5)$ is not: the fraction $x/y$ is not in $k[x,y]/(x^2-y^5)$ but is a zero of the monic polynomial $f(s)=s^2 -y^3$.

share|improve this answer
add comment

We can take advantage that the ring $k[t]$ is a UFD. It seems to me that we don't require $k$ to be algebraically closed, so I shall drop that assumption.

Assume contrariwise that there exists an isomorphism $f:k[x,y]/(x^2-y^5)\to k[t]$. Let $P(t)=f(x)$ and $Q(t)=f(y)$ be the respective images of $x$ and $y$. Using the factorization we can write $$ P(t)=a\prod_ip_i(t)^{n_i}\qquad\text{and}\qquad Q(t)=b\prod_jq_j(t)^{m_j}, $$ for some constants $a,b\in k$ and some irreducible monic polynomials $p_i,q_j\in k[t]$. We must have $P(t)^2=f(x^2)=f(y^5)=Q(t)^5$, so by the uniqueness of factorization we get (after the usual renumbering) $a^2=b^5$, $p_i(t)=q_i(t)$, and $2n_i=5m_i$ for all $i$. So all the exponents $n_i$ are divisible by five, and all the exponents $m_i$ are even. Write $u=\prod_ip_i(t)^{n_i/5}\in k[t].$ We have $$ P=a\cdot u^5,\qquad Q=b\cdot u^2,\qquad\text{and}\qquad a^2=b^5. $$ We see that the image of $f$ is the subring $R=k[u^2,u^5]$ of $k[t]$. Clearly $$R=\bigoplus\sum_{i\in\mathbf{N},i\notin\{1,3\}}k\cdot u^i$$ is a proper subring of $k[u]$. As $k[u]$ is a subring of $k[t]$ (these two rings could possibly be equal), we see that $f$ cannot be surjective.

share|improve this answer
4  
+1: This is a very nice proof since it is completely elementary in a logical sense, albeit complicated. By contrast it illustrates how introducing more sophisticated concepts like smoothness and integrality simplifies proofs. There is an interesting moral for beginners here: abstraction is for your good and not only for the pleasure of sadistic professors . –  Georges Elencwajg Mar 28 '12 at 8:40
    
Everything that you say is very true, @Georges :-) OTOH I also enjoy the kind of exercises, where elementary methods are taken to their limits. After all, when (or, "if") we are pushing the limits of our understanding further out, we also need to know how to milk every last ounce out of the tools we have. Granted, the first year calculus students don't necessarily appreciate the kind of exercise, where they have to prove that $f(x)=x^3$ is an increasing function WITHOUT using the derivative. –  Jyrki Lahtonen Mar 28 '12 at 9:21
add comment

Below is a very simple proof accessible at high-school level.

Theorem $\ $ If $\rm\: a^2 = b^5\:$ in UFD $\rm\:R[x]\:$ then $\rm\:R[a,b]\:$ is a proper subring of $\rm\:R[x]\:$

Proof $\ $ Clear if $\rm\:a,b\in R.\:$ Else $\rm\:a\:$ or $\rm\:b\not\in R,\:$ so one is a nonunit, so has a prime factor $\rm\:p\not\in R.\:$ By $\rm\:a^2 = b^5,\:$ prime $\rm\:p\ |\ a\!\iff\! p\ |\ b,\:$ so $\rm\:p\ |\ a,b.\:$ Comparing powers of $\rm\:p\:$ in $\rm\:a^2 = b^5\:$ shows $\rm\:p^2\:|\ a,b.\:$ Thus $\rm\:p\not\in R[a,b]\:$ else $\rm\:p = r + a f + b g,\:$ $\rm r\in R\:$ so $\rm\:p^2\:|\:a,b\:$ $\Rightarrow$ $\rm\:p\ |\ r\:$ $\Rightarrow$ $\rm\:r=0\:$ $\Rightarrow$ $\rm\:p^2\ |\ p.$

share|improve this answer
add comment

This is a minor (and nitpicking) complement to Georges and Sebastian's answers. (I up-voted the question and all answers.) I would state things as follows. Let me change the notation to $$ K[X,Y],\quad K[x,y]:=K[X,Y]/(X^2-Y^5). $$ Assume by contradiction that $K[x,y]$ is a UFD.

The square of $x/y$ being $y^3$, we see that $y$ divides $x$.

Any element of $K[x,y]$ can be written in a unique way as $p(y)+x\,q(y)$ with $p,q\in K[Y]$. Thus we have $$ x=y\ (p(y)+x\ q(y))=y\ p(y)+ x\ (y\ q(y)), $$ which implies $p(Y)=0$ and $Yq(Y)=1$, contradiction.

share|improve this answer
    
I don't understand "we see that $y$ divides $x$". –  Martin Brandenburg Mar 28 '12 at 10:12
1  
Dear @Martin: Assume that $K[x,y]$ is a UFD, write $x/y=ua_1^{n_1}\cdots a_k^{k_1}$, with $u$ a unit, the $a_i$ non-associated irreducibles, $n_i\in\mathbb Z$, plug this into $(x/y)^2=y^3$, and note that $n_i\ge0$ for all $i$. –  Pierre-Yves Gaillard Mar 28 '12 at 10:38
    
Alright this is just UFD => normal. –  Martin Brandenburg Mar 29 '12 at 9:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.